mreditor16
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Sorry I should have clarified. I got the first part out, and I am stuck on the 11th vector part. According to the answers, there is a 11th vector :/10 vectors with components all 0 except for one which is 5, so 5e1, 5e2, etc where ei are the standard basis vectors in R^10. You can then see that there is no 11th vector that satisfies the given conditions.
Your working seems correct, but I think OP would also want a justification as to why each component of the 11th vector need to be equal, otherwise you wouldn't really be getting the same distance for EVERY pair. To make things simpler, try the exact same case for R^2 and R^3 (where you would look for a 3rd and 4th vector respectively) and the same method would apply for R^10.Question OP, why doesn't your tutor help you during tutorials?
Not sure if my solution is valid but VBN clarifies things better.
Vector a = (β,β,...β) ε R^10
Find β such that
|ai - a| = 5sqrt2
ai - a = (-β,..-β,α-β,-β,-β)
|ai - a|= 9β^2 +(α-β)^2 = 50
9β^2 +(5-β)^2 = 50
Then solve for β or whatever you called the parameter.
My tutor's 1 minute explanation didn't make sense at all. Hence, me asking it on BOS.Question OP, why doesn't your tutor help you during tutorials?
Not sure if my solution is valid but VBN clarifies things better.
Vector a = (β,β,...β) ε R^10
Find β such that
|ai - a| = 5sqrt2
ai - a = (-β,..-β,α-β,-β,-β)
|ai - a|= 9β^2 +(α-β)^2 = 50
9β^2 +(5-β)^2 = 50
Then solve for β or whatever you called the parameter.
Do you remember what your tutor said?My tutor's 1 minute explanation didn't make sense at all. Hence, me asking it on BOS.
Tbh not much, because I think we only had a few minutes left and they miraculously seemed to pluck the quadratic out of nowhere, and how they got said quadratic was really confusing.Do you remember what your tutor said?
Fair enough then, I asked because then maybe I could have reasoned out what they meant and tried to deduce some easier way of solving this, especially if it took them around a minute to do (unless they knew the q beforehand, and pulled the quadratic of out nowhere). Wonder if anyone else can shed some light on an easier method for this question? That would be good.Tbh not much, because I think we only had a few minutes left and they miraculously seemed to pluck the quadratic out of nowhere, and how they got said quadratic was really confusing.
But how would you know all components need to be the same? Why can't they be different?Another method is just substitute in the ansatz (where there are 10 a's), and solve for a and hope you get a real solution (which you do in the end), rather than proving you need all components same.
Fair enoughI didn't say they can't be different. You could just seek one that has all same components and see if you can solve the resulting equation.